Turing Machine Resisting Isolated Bursts of Faults

Ilir Capuni and Peter GacsBoston University

Introduction

‘Regular’ transition A faulty transition

A fault: a violation of the transition function

Change the state, the tape symbol and move the head arbitrarily

0 0 1 0 …… 1

q

M

0 1 0 …… 0

q’

M

0 0 0 …… 1

q’

1

0step i

step i+1

q

)1,1,'()1,( qq

Is there a Turing machine that can carry out its computation despite faults (violations of the transition function) that occur independently of each other with small probability?

The main question

Reduced goal: resistance to isolated bursts of faults

0 0 1 0 …… 1

q

M

)(fV

1 0 0 1 0 0

Our result illustrated

0 0 0 0 1 0 1 0 0 0 0 …… 1

a3 a1 a2 a3 b1 b2… b3 …

3) Block code (E , D) of block-size Q

a1 a2 a3 b1 b2 b3 a1 a2 a3

Given:

2M

1) Integers V, Q, C, all depending linearly on

We construct:

1M

q

)|,(||| 21 poly

)|,(||| 21 poly

2) Machine with the alphabet and state set that does not contain a halting state

11

1) Machine with the alphabet and state set 22

2) The size of the bursts 0

Such that

0 0 0 0 1 0 1 0 0 0 0 …… 1

0 0 0 0 1 0 1 0 0 0 0 …… 1

qf

q5

0 0 0 0 1 0 1 0 0 0 0 …… 0 q0

…

0

M2

1

T-1

T

0 …… 1

a a a a b b b a a a a …… b

a a a a b b b ……

… M1

t-Q

t, t>CT

No faults for Q steps

qf

q’… …pt

p5

p0

Burst are separated by at

least V fault- free steps

The result

Sketch of the construction

0 0 1 0 …… 1

q

0 0 0 0 0 1 1 1 1q q q q

0 1 2 3 4 1 2 3 4L L L L L 2 2 2 2

_ _ _ _ _

1 1 1 11 1 1 1 1

1 1 1 1 1

0 1 2 3 4g g g g g1 1 1 1 1

_ _ _ _ _InfoStateAddrSweepDrift

Other fields

1q

021

Head

Mode = NormalAddr = 0Sweep = 2Other fields

Colonies

E

1M

2M… …2

0 0 0 0 0 1 1 1 1q q q q

0 1 2 3 4 1 2 3 4L L L L L L L L L

_ _ _ _ _

1 1 1 11 1 1 1 1

1 1 1 1 1

0 1 2 3 4g g g g g1 1 1 1 1

_ _ _ _ _InfoStateAddrSweepDrift

Other fields

Head

Mode = NormalAddr = 0Sweep = 1Other fields

… …

Com

puta

tion

ph

ase

Transf

err

ing

ph

ase

Two lines of defense

Information: there is an error-correcting code and global repetitions to handle computation errors

Simulation: There is a structure that could be restored locally using a “recovery procedure” that does not restore lost information

Burst -tolerant Tur(n)ing machine

Intervals of cells whose structure is altered by faults will be called islands

The cells where Info and State are altered by faults will be called stains

Islands and stains

An example

Recovery procedure

Simulation resumes, but a stain can remain until the next decoding-encoding takes place over this colony

Localization with zigging

Extensive damage Damage localized with zigging

Sweep + 2 Sweep - 1 Sweep + 1 Sweep - 1

Alarm is called

1822 front

front

Opens a constant size recovery interval around the cell where the Alarm is called

In several sweeps it computes majority of Address, Sweep, and Drift fields, and computes the values that need to be filled in the ‘damaged’ area

Then it leaves the head close to the front of the sweep

Recovery procedure

Disturbing the recovery

Bad news Good news

The burst can leave the machine in some arbitrary state

The recovery procedure can itself be also disturbed by a burst

Our recovery procedure handles provably all these cases

Certain proofs are long and require detailed case analysis

Proofs

Coding: Store the program and the state of M2 on the tape of M1 and make M1 act as a Universal Turing machine

The slowdown of the machine now becomes quadratic on the burst size

Extensions

0 0 1 0 …… 1

q

1M2M

…

a a b b b b a b aa b a b

0 1 2 3 4 1 2 3 4L L L L L 2 2 2 2

_ _ _ _ _

1 1 1 11 1 1 1 1

a b a b a

0 1 2 3 4g g g g g1 1 1 1 1

_ _ _ _ _InfoStateAddrSweepDrift

Other fields

ab

021

Head

Mode = NormalAddr = 0Sweep = 2Other fields

…2

0 1 a 1 0 b 1 0 a 0 1 a 1 0a Prog

Our noise model was combinatorial How to deal with low-probability noise

combinatorially?◦ Level 1: Consider first noise that has low frequency:

Bursts are b1 long but V1 steps apart from each other

◦ Level 2: Then allow violations of the previous occurring with low frequency:

Occur in at most b2 steps separated by at least V2 steps from each other

◦ And so on…

We are left with a noise set that is “sparse”

Toward resistance to probabilistic noise

We use the previous construction as a building block

Some serious additional challenges must be handled to maintain the hierarchy of simulations

From the Tur(n)ing machine to a Turing machine that can withstand probabilistic noise

1M 2M 3M kM M…

0 0 1 0 …… 1

q

3MM

…

a a b b b b a b aa b b b

0 1 2 3 4 1 2 3 4L L L L L 2 2 2 2

_ _ _ _ _

1 1 1 11 1 1 1 1

a b a b a

0 1 2 3 4g g g g g1 1 1 1 1

_ _ _ _ _InfoStateAddrSweepDrift

Other fields

aa

021

Head

Mode = NormalAddr = 0Sweep = 2Other fields

…2

0 1 a 1 0 1 a 1 0 0 1 a 1 00 Prog

a a b b b b a b aa c b b

0 1 2 3 4 1 2 3 4L L L L L 2 2 2 2

_ _ _ _ _

1 1 1 11 1 1 1 1

a b a b a

0 1 2 3 4g g g g g1 1 1 1 1

_ _ _ _ _InfoStateAddrSweepDrift

Other fields

aa

021

Head

Mode = NormalAddr = 0Sweep = 2Other fields

2M… …2

0 1 a 1 0 1 a 1 0 0 1 a 1 00 Proga a b b b b 0 b a

1 c b b

0 1 2 3 4 1 2 3 4L L L L L 2 2 2 2

_ _ _ _ _

1 1 1 11 1 1 1 1

a b a b a

0 1 2 3 4g g g g g1 1 1 1 1

_ _ _ _ _InfoStateAddrSweepDrift

Other fields

1a

021

Head

Mode = NormalAddr = 0Sweep = 2Other fields

1M

… …2

0 1 a 1 0 1 a 1 0 0 1 a 1 00 Prog

T

It is a natural simple question Just as with cellular automata, it is surprising that

the solutions seem to require so much complexity So far, this is the simplest universal machine that

can resist isolated bursts of faults

Why do we care?

Q u s t i… e

q

o n s ?

q q q q P q’ q’’ f f

h a n k _ y o u !

f l k k k k k k k k